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Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?
Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?
A. Zero B. One C. Two D. Three E. More than three
To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.
Three values that would make the set symmetrical are 0, 10, and 20:
{0, 4, 8, 12, 16}
{4, 8, 10, 12, 16}
{4, 8, 12, 16, 20}
We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).
Well, we can set up three scenarios, each with a relevant equation.
(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median: \(\frac{40 + x}{5} = 8\) \(40 + x = 40\) \(x = 0\)
(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median: \(\frac{40 + x}{5} = x\) \(40 + x = 5x\) \(40 = 4x\) \(x = 10\)
(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median: \(\frac{40 + x}{5} = 12\) \(40 + x = 60\) \(x = 20\)
We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.
Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?
A. Zero B. One C. Two D. Three E. More than three
To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.
Three values that would make the set symmetrical are 0, 10, and 20:
{0, 4, 8, 12, 16}
{4, 8, 10, 12, 16}
{4, 8, 12, 16, 20}
We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).
Well, we can set up three scenarios, each with a relevant equation.
(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median: \(\frac{40 + x}{5} = 8\) \(40 + x = 40\) \(x = 0\)
(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median: \(\frac{40 + x}{5} = x\) \(40 + x = 5x\) \(40 = 4x\) \(x = 10\)
(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median: \(\frac{40 + x}{5} = 12\) \(40 + x = 60\) \(x = 20\)
We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.
Answer: D
Hi,
Isnt it necessary for the values to be in AP,for the mean and median to be equal?
Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?
A. Zero B. One C. Two D. Three E. More than three
To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.
Three values that would make the set symmetrical are 0, 10, and 20:
{0, 4, 8, 12, 16}
{4, 8, 10, 12, 16}
{4, 8, 12, 16, 20}
We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).
Well, we can set up three scenarios, each with a relevant equation.
(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median: \(\frac{40 + x}{5} = 8\) \(40 + x = 40\) \(x = 0\)
(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median: \(\frac{40 + x}{5} = x\) \(40 + x = 5x\) \(40 = 4x\) \(x = 10\)
(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median: \(\frac{40 + x}{5} = 12\) \(40 + x = 60\) \(x = 20\)
We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.
Answer: D
Hi,
Isnt it necessary for the values to be in AP,for the mean and median to be equal?
___________________ No, consider {1, 1, 2, 2}.
_________________
Hi, It does not say anywhere, that the value of x should be positive or equal to zero. This bring a confusion and reflects to the answer.
Kind regards
The question asks about all possible values of x, negative, 0, positive, for which he mean of set S will be equal the median of set S. There are only three values possible: 0, 10 and 20.
_________________
Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?
A. Zero B. One C. Two D. Three E. More than three
To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.
Three values that would make the set symmetrical are 0, 10, and 20:
{0, 4, 8, 12, 16}
{4, 8, 10, 12, 16}
{4, 8, 12, 16, 20}
We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).
Well, we can set up three scenarios, each with a relevant equation.
(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median: \(\frac{40 + x}{5} = 8\) \(40 + x = 40\) \(x = 0\)
(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median: \(\frac{40 + x}{5} = x\) \(40 + x = 5x\) \(40 = 4x\) \(x = 10\)
(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median: \(\frac{40 + x}{5} = 12\) \(40 + x = 60\) \(x = 20\)
We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.
Answer: D
Hello, It says that the numbers need not be ascending order. So, why aren't we taking the case {4, 8, 16, 12, x} ? which implies x= 40. Therefore, values of x is more than three. Thanks in advance!
Set S consists of 5 values, not necessarily in ascending order: {4, 8, 12, 16, \(x\)}. For how many values of \(x\) does the mean of set S equal the median of set \(S\)?
A. Zero B. One C. Two D. Three E. More than three
To solve this problem quickly, you might try to come up with likely values for \(x\) that would make the mean equal the median. One sort of set for which the mean equals the median is a set with values symmetrically spaced around its mean/median. The values do not have to be evenly spaced.
Three values that would make the set symmetrical are 0, 10, and 20:
{0, 4, 8, 12, 16}
{4, 8, 10, 12, 16}
{4, 8, 12, 16, 20}
We are down to choices (D) and (E). Now, can we prove that no other values of \(x\) make the mean equal the median? After all, some non-symmetrical sets have their mean equal to their median: for instance, {1, 1, 2, 2.5, 3.5}. All you need to do is make the "residuals," or differences, around the middle value cancel out (in the case above, the values to the left of 2 are 1 & 1, leaving a total residual of -2, while the values to the right of 2 are 2.5 and 3.5, leaving a total residual of +2).
Well, we can set up three scenarios, each with a relevant equation.
(1) If \(x\) is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median: \(\frac{40 + x}{5} = 8\) \(40 + x = 40\) \(x = 0\)
(2) If \(x\) is between 8 and 12, then the median is equal to \(x\). Again, we set the mean equal to the median: \(\frac{40 + x}{5} = x\) \(40 + x = 5x\) \(40 = 4x\) \(x = 10\)
(3) If \(x\) is greater than 12, then the median is equal to 12. Again, we set the mean equal to the median: \(\frac{40 + x}{5} = 12\) \(40 + x = 60\) \(x = 20\)
We have now exhausted all the possibilities for \(x\). In fact, we did not have to actually compute the values of \(x\) in each case; rather, we could have simply realized that each equation is linear in \(x\) and so would have exactly one solution. Since there are three scenarios, there are exactly three values of \(x\) that satisfy the constraint of making the mean and the median equal. Indeed, if we had started with this approach, we might have gotten to the answer more quickly.
Answer: D
Hello, It says that the numbers need not be ascending order. So, why aren't we taking the case {4, 8, 16, 12, x} ? which implies x= 40. Therefore, values of x is more than three. Thanks in advance!
In order to compute the median, values have to be in order (either ascending or descending). Therefore your statement is fallacious.
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38% (01:38) correct 
